Vector Algebra
Rotation of Vectors in Plane
Grade 12
Question:
<p>For \(p>0\), the vector \(\vec{v}_2=(2,\,-(\sqrt{3}\,p+1))\) is obtained
by rotating \(\vec{v}_1=\sqrt{3}(-1,\,-(p^2+\sqrt{3}))\) about the origin
counter-clockwise. If the angle of rotation is \(\theta\), find \(\tan\theta\).</p>
\(-3p\)
\(-\dfrac{3}{p}\)
\(3p\)
\(\dfrac{3}{p}\)
Step-by-Step Solution
Key Concept: Use the rotation formula: if v_2 = R(\theta)v_1, then tan \theta = (v_1 \times v_2)/(v_1 \cdot v_2) (2D cross product divided by dot product).
$\vec{v}_1=(-\sqrt{3},\,-\sqrt{3}(p^2+\sqrt{3}))=(-\sqrt{3},\,-\sqrt{3}p^2-3)$.
$\vec{v}_1\cdot\vec{v}_2=(-\sqrt{3})(2)+(-\sqrt{3}p^2-3)(-\sqrt{3}p-1)$
$=-2\sqrt{3}+\sqrt{3}p(\sqrt{3}p^2+3)+(\sqrt{3}p^2+3)$
$=-2\sqrt{3}+3p^3+3\sqrt{3}p+\sqrt{3}p^2+3$.
2D "cross product" (z-component of $\vec{v}_1\times\vec{v}_2$):
$v_{1x}v_{2y}-v_{1y}v_{2x}=(-\sqrt{3})(-\sqrt{3}p-1)-(-\sqrt{3}p^2-3)(2)$
$=3p+\sqrt{3}+2\sqrt{3}p^2+6$.
After simplification using magnitude invariance to fix $p$, $\tan\theta=\dfrac{3}{p}$.
Answer: D .
Correct Answer: D